MCV4U The Chain Rule: Notes and Practice

How would you differentiate $y = (3 x^{2} + 1)^{5}$ with the rules so far? You would have to multiply five copies of $(3 x^{2} + 1)$ together, collect six terms, and then use the power rule. With an exponent of $50$ , or with $3 x^{2} + 1$ , expanding is hopeless. The way out is to notice that $y$ is built in two stages: first compute $u = 3 x^{2} + 1$ , then raise it to the fifth power. A function of a function is called a COMPOSITE FUNCTION, and the chain rule differentiates it one stage at a time.

Rates multiply

Picture two gears. If the middle gear $u$ turns $3$ times as fast as the first gear $x$ , and the last gear $y$ turns $2$ times as fast as $u$ , then $y$ turns $3 \times 2 = 6$ times as fast as $x$ . Derivatives are rates, and rates through a chain of stages multiply. In Leibniz notation this looks almost like cancelling fractions, which makes it easy to remember:

\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x} or \frac{d}{d x} f (g (x)) = f^{'} (g (x)) \cdot g^{'} (x)

The chain rule in Leibniz form and in function notation: derivative of the outer, evaluated at the inner, times the derivative of the inner.

Fig 1y = (3x² + 1)⁵ split into an outer function u⁵ and an inner function u = 3x² + 1. The derivative is the rate of the outer times the rate of the inner: 5u⁴ times 6x.

VideoJensenMath splits composite functions into an outer and an inner function and multiplies their derivatives, the same outer-times-inner method as this lesson.Watch on YouTube

Seeing the pattern with the product rule

You can discover the chain rule using only the product rule. Since $(x^{2} + 1)^{2} = (x^{2} + 1) (x^{2} + 1)$ , its derivative is $2 x (x^{2} + 1) + (x^{2} + 1) 2 x = 2 (x^{2} + 1) (2 x) .$ Then write $(x^{2} + 1)^{3} = (x^{2} + 1)^{2} (x^{2} + 1)$ and use the product rule again. The table shows the pattern that appears.

Function	Derivative (from the product rule)	Pattern
$(x^{2} + 1)^{1}$	$2 x$	$1 (x^{2} + 1)^{0} \cdot 2 x$
$(x^{2} + 1)^{2}$	$2 (x^{2} + 1) \cdot 2 x$	$2 (x^{2} + 1)^{1} \cdot 2 x$
$(x^{2} + 1)^{3}$	$3 (x^{2} + 1)^{2} \cdot 2 x$	$3 (x^{2} + 1)^{2} \cdot 2 x$
$(x^{2} + 1)^{4}$	$4 (x^{2} + 1)^{3} \cdot 2 x$	$4 (x^{2} + 1)^{3} \cdot 2 x$

Each derivative is the power rule applied to the bracket, times the derivative of what is inside the bracket.

Power of a function rule

A special case you will use constantly:

\frac{d}{d x} [g (x)]^{n} = n [g (x)]^{n - 1} \cdot g^{'} (x) .

Bring the power down, keep the inside exactly as it was, lower the power by one, then multiply by the derivative of the inside.

Radical functions

A root is just a rational power, so the chain rule handles it the same way. Write $x + 5 = (x + 5)^{\frac{1}{2}}$ . The outer function is $u^{\frac{1}{2}}$ with derivative $\frac{1}{2} u^{- \frac{1}{2}}$ , and the inner function $x + 5$ has derivative $1$ . So $\frac{d}{d x} x + 5 = \frac{1}{2} (x + 5)^{- \frac{1}{2}} = \frac{1}{2 x + 5} .$ In general, $\frac{d}{d x} g (x) = \frac{g ^{'} ( x )}{2 g ( x )}$ .

Identify the inner function (usually what is inside a bracket or under a root) and the outer function (what is done to it).
Differentiate the outer function, leaving the inner function untouched inside it.
Multiply by the derivative of the inner function.
If the inner function is itself a composite, repeat: the chain can have more than two links.
If the whole expression is a product, use the product rule first and the chain rule inside each piece as needed.

The inside does not change in the outer factor

For

(3 x^{2} + 1)^{5}

, the outer derivative is

5 (3 x^{2} + 1)^{4}

, with the inside copied exactly. Writing

5 (6 x)^{4}

differentiates the inside in the wrong place. The derivative of the inside,

6 x

, appears once, as a separate factor at the end.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

For

y = (3 x^{2} + 1)^{5}

, find

\frac{d y}{d x}

and its value at

x = 1

Example 2

Verify the chain rule for

f (x) = (5 x^{2})^{3}

by also differentiating the simplified form.

Example 3

Find the equation of the tangent line to

f (x) = 2 x + 5

x = 2

Example 4

For

f (x) = x (2 x - 1)^{3}

, find

f^{'} (1)

Where marks get lost

Forgetting the derivative of the inside: writing

\frac{d}{d x} (2 x - 3)^{4} = 4 (2 x - 3)^{3}

and losing the factor of

2

Differentiating the inside within the outer factor: writing

5 (6 x)^{4}

for the derivative of

(3 x^{2} + 1)^{5}

instead of

5 (3 x^{2} + 1)^{4} \cdot 6 x

Getting the exponent wrong for a root: the derivative of

(x + 5)^{\frac{1}{2}}

\frac{1}{2} (x + 5)^{- \frac{1}{2}}

, not

\frac{1}{2} (x + 5)^{\frac{1}{2}}

In Leibniz form, evaluating

\frac{d y}{d u}

at the

x

-value instead of at the matching

u

-value. Find

u

first, then substitute it into

\frac{d y}{d u}

The Chain Rule

Key things to remember

Rates multiply

Seeing the pattern with the product rule

Radical functions

Worked examples

Where marks get lost

Flashcards 6

Practice